Let $g[n]=1$ for n = 0 and zero otherwise, and $h[n]=1$ for n = 0,3,6,9,… and zero otherwise. If $y[n]=h[n]\otimes g[n]$ with convolution denoted by ⊗, find y[2].
The sequence g[n] is the unit impulse δ[n]. Convolution with the unit impulse leaves the other sequence unchanged: $h[n]*\delta[n]=h[n]$. Since 2 is not a nonnegative multiple of 3, h[2] = 0. Therefore y[2] = 0.
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